Chemistry combustion analysis gas problem

Let's try this again.In summary, the complete combustion of a 0.0150 mol sample of a hydrocarbon, CxHy, results in 1.680L of CO2 at STP and 0.810g of H2O. The molecular formula is most likely C5H6, but it could also be C10H12 or C15H18. It is not possible to determine the exact molecular formula without more information. The empirical formula is C4H5.
  • #1
Nellen2222
55
0

Homework Statement



Complete combustion of a 0.0150 mol sample of a hydrocarbon, CxHy, gives 1.680L of CO2 at STP and 0.810g of H2O
a) what is the molecular formula
b) what is empirical formula

Homework Equations



pv=nRt

The Attempt at a Solution



a) mol of co2: n=pv/Rt=1 atm * 1.680L/0.0821*273= 0.074955495mol

- Mol of h2o: 0.810g/18.02g/mol * 2 = 0.08990011 mol H2

mol ratio: 0.08990011/0.074955495= 1.199379845... * 5 = 6

molecular formula: c5h6.

Is this wrong? is it possible to multiply by four and round the number to 5 instead of 6? so that it would be c4h5? Or is the above one correct?also: how do i get the molecular formula? i can't get a lowest whole number ratio here.
Thanks!
 
Physics news on Phys.org
  • #2
Looks good to me provided you did the math correctly. Could also be C10H12 or C15H18...
 
Last edited:
  • #3
Nellen2222 said:
0.0150 mol sample of a hydrocarbon

chemisttree said:
Could also be C10H12 or C15H18...

C5H6 it is, OP started with the number of moles.
 
  • #4
Argh! That should have been Nellen's reply.
 
  • #5
Oops.
 

Related to Chemistry combustion analysis gas problem

What is combustion analysis in chemistry?

Combustion analysis in chemistry is a method used to determine the chemical composition of a substance by burning it in the presence of oxygen. This process produces combustion products that can be analyzed to determine the relative amounts of carbon, hydrogen, and oxygen present in the substance.

What is the purpose of conducting a combustion analysis?

The purpose of conducting a combustion analysis is to determine the empirical formula or molecular formula of a substance. This information can be used to identify the substance, as well as to calculate other important properties such as percentage composition and molar mass.

What are the steps involved in a combustion analysis?

The steps involved in a combustion analysis include accurately weighing a sample of the substance, burning it in the presence of excess oxygen, collecting the combustion products, and analyzing them using techniques such as gas chromatography or mass spectrometry. The data obtained from the analysis can then be used to calculate the empirical formula or molecular formula of the substance.

What is the gas problem in combustion analysis?

The gas problem in combustion analysis refers to the presence of other gases in the combustion products that can interfere with the accurate determination of the carbon, hydrogen, and oxygen content. These gases can include water vapor, nitrogen, and sulfur dioxide. To solve this problem, the combustion products must be carefully purified before analysis.

What are some potential sources of error in combustion analysis?

Some potential sources of error in combustion analysis include incomplete combustion, contamination of the sample, and errors in measurement. Incomplete combustion can lead to inaccurate results, while contamination of the sample can introduce other elements that can affect the analysis. It is important to carefully control and monitor all variables in order to minimize errors in combustion analysis.

Similar threads

  • Biology and Chemistry Homework Help
Replies
1
Views
2K
  • Biology and Chemistry Homework Help
Replies
5
Views
4K
  • Biology and Chemistry Homework Help
Replies
1
Views
2K
  • Biology and Chemistry Homework Help
Replies
7
Views
2K
  • Biology and Chemistry Homework Help
Replies
4
Views
6K
  • Biology and Chemistry Homework Help
Replies
2
Views
6K
  • Biology and Chemistry Homework Help
Replies
4
Views
2K
  • Biology and Chemistry Homework Help
Replies
6
Views
6K
  • Biology and Chemistry Homework Help
Replies
7
Views
7K
  • Biology and Chemistry Homework Help
Replies
5
Views
15K
Back
Top